URGENT: Differential calculus

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December 14th 2007, 01:29 PM
Rydbirk

URGENT: Differential calculus

Okay, my english is not perfect, but I hope you can understand what I'm talking about.

I'm having this subject about movement and air resistance. We're watching an object dropping in an area with air resistance. The resulting force on the object is (using Newton's 2. law):
F_res = -F_grav + F_air <=> ma = -mg + k*v^2

Let's introduce s(t) (don't know what it's called in english), and thereby s'(t) = v(t) and s''(t) = a(t).

After some paraphrasing, we get a(t) = k/m * v^2 - g <=> s''(t) = C * (s'(t))^2 - g with C = k/m

To find v(t) and s(t) we integrate, so
v(t) = 1/3 * C * t * (s(t))^3 - g * t + v_0 and
s(t) = 1/12 * C * t^2 * ((int.)s(t)dt)^4 - g * t^2 + v_0*t + s_0

Is this correct? How do I find s(t)??? My calculator can't find it.. Maybe I'm typing something wrong or I've done something wrong.. It's really urgent, please help!
December 14th 2007, 03:27 PM
Rydbirk

D'oh!

I made a mistake! You can't integrate like that..

$s'(t)=\int{(s'(t))^2dt}*C-g*t+v_{0}$
That makes it easier. Well, this is my first time using seperation ^^ so I might return..
December 14th 2007, 04:34 PM
topsquark

Quote:

Originally Posted by Rydbirk

Okay, my english is not perfect, but I hope you can understand what I'm talking about.

I'm having this subject about movement and air resistance. We're watching an object dropping in an area with air resistance. The resulting force on the object is (using Newton's 2. law):
F_res = -F_grav + F_air <=> ma = -mg + k*v^2

Let's introduce s(t) (don't know what it's called in english), and thereby s'(t) = v(t) and s''(t) = a(t).

After some paraphrasing, we get a(t) = k/m * v^2 - g <=> s''(t) = C * (s'(t))^2 - g with C = k/m

To find v(t) and s(t) we integrate, so
v(t) = 1/3 * C * t * (s(t))^3 - g * t + v_0 and
s(t) = 1/12 * C * t^2 * ((int.)s(t)dt)^4 - g * t^2 + v_0*t + s_0

Is this correct? How do I find s(t)??? My calculator can't find it.. Maybe I'm typing something wrong or I've done something wrong.. It's really urgent, please help!

I think your English is pretty good. (Handshake)

s(t) is called the "displacement." It roughly corresponds to "position."

No, your work is not correct. You have a differential equation
$s''(t) = C * (s'(t))^2 - g$

But
$\int (s'(t))^2~dt \neq \frac{1}{3}(s'(t))^3$

You need to first write this as a velocity equation:
$v'(t) = C(v(t))^2 - g$

At the moment I can't remember how to solve this problem. Perhaps one of the other Helpers will post on this.

-Dan
December 14th 2007, 04:42 PM
Rydbirk

Yes

Yes, I've concluded that as well - but thanks ;) And thank you for the information.

But now I'm stuck again. I've spent 2-3 hours working with this and I've recieved a lot of help, but I haven't been tought how to deal with separation in differential calculations - so I beg you, please, somebody, show me how to solve the following step-by-step and explain to me, how you do. Please, it's very important!

$s''(t) = C*(s'(t))^2-g$
December 14th 2007, 04:59 PM
topsquark

Quote:

Originally Posted by topsquark

$v'(t) = C(v(t))^2 - g$

I'm an idiot. I just remembered how to do this.

$\frac{dv}{dt} = Cv^2 - g$

$\frac{dv}{Cv^2 - g} = dt$

$\int \frac{dv}{Cv^2 - g} = \int~dt$

$\frac{1}{2\sqrt{Cg}}~ln \left ( \frac{v\sqrt{C} - \sqrt{g}}{v\sqrt{C} + \sqrt{g}} \right ) = t + A$
where A is the integration constant, set by the initial conditions.

As you can see this is not a nice equation to solve for v, but it can be done:
$v(t) = -\sqrt{\frac{g}{C}}~\frac{exp[2t \sqrt{Cg} + 2A \sqrt{Cg}] + 1}{exp[2t\sqrt{Cg} + 2A\sqrt{Cg}] - 1}$

I'll leave it to you to integrate this to get s(t). (This is even less pretty, but still doable.)

-Dan
December 14th 2007, 05:13 PM
Rydbirk

Thanks!!

Topsquark, thank you thank you thank you!!

So A must be my $v_{0}$ .

There just 2 steps I don't get, though..
How do you isolate $\frac{1}{2\sqrt{Cg}}$ ? And how do you go from (4) to (5) ?

If noone has the time to help me out, I'll return in the morning - in Denmark it's 3 AM ;)
December 14th 2007, 05:25 PM
topsquark

Quote:

Originally Posted by Rydbirk

Topsquark, thank you thank you thank you!!

So A must be my $v_{0}$ .

There just 2 steps I don't get, though..
How do you isolate $\frac{1}{2\sqrt{Cg}}$ ? And how do you go from (4) to (5) ?

If noone has the time to help me out, I'll return in the morning - in Denmark it's 3 AM ;)

$\int \frac{dx}{x^2 - a^2} = \int \left ( \frac{A}{x + a} + \frac{B}{x - a} \right )~dx$
using partial fractions. Find A and B and you can integrate term by term.

The constant out in the front can be found by changing the integration variable from v to $u = \sqrt{C} \cdot v$ . The rest of the constant comes from the partial fraction coefficients A and B.

-Dan
December 17th 2007, 04:37 AM
Rydbirk

New thread

See this thread for further help:
http://www.mathhelpforum.com/math-he...-calculus.html

:o
December 18th 2007, 01:37 PM
Rydbirk

Hmmm...

My calculator gives me a different solution:

$<br /> \frac{1}{2\sqrt{Cg}}~ln \left ( \frac{\sqrt{g} + v\sqrt{C}}{\sqrt{g} - v\sqrt{C}} \right ) = t + A<br />$

But why?